Friday

October 31, 2014

October 31, 2014

Posted by **John** on Sunday, March 4, 2012 at 2:31pm.

bounded by the curves y = x^2 and y = mx is equal to 8.

I would very much appreciate it if someone could find the answer and explain how you did it.

- calculus -
**Abby**, Wednesday, March 6, 2013 at 12:25amthe anti derivative is (m/2)x^2 - 1/3(x)^3 so if you use the 2nd fundamental theorem of calculus using (mx - x^2) over the interval (0,m) and set that equal to 8 (the interval is 0 to m because the function with the larger area is mx) you should get m^3 / 8 = 8. Solve for m and you're done :)

**Answer this Question**

**Related Questions**

calculus - Consider the graphs of y = 3x + c and y^2 = 6x, where c is a real ...

calculus - use the methods of this section to sketch the curve. y=x^3-3a^2x+2a^3...

math-calculus 2 - Consider the given curves to do the following. 64 y = x^3, y...

Calculus-Area between curves - Sketch the region enclosed by the given curves. ...

CALCULUS - Sketch the region enclosed by the given curves. y = 4/X y = 16x, y...

Calculus Area between curves - Sketch the region enclosed by the given curves. ...

calculus - Determine the exact value for the constant k such that the area of ...

calculus - consider the area enclosed between the curves f(x)=x^2 and g(x)=4x ...

calculus - consider the area enclosed between the curves f (x) = x2 and g (x) = ...

Physics - A car travels the same distance at constant speed around two curves, ...